Petroff equation for bearing friction

Petroff's law

Petroff assumed a lightly loaded journal running concentric with the bearing, with the clearance $c$ filled with oil, so that the shear of the oil film gives the friction. The law states that the friction torque is proportional to viscosity and speed and inversely proportional to the clearance.

Derivation

Journal radius $r$, length $l$, speed $N$ rev/s, load $W$.

Surface velocity $U = 2\pi r N$. Shear stress in the film $\tau = \mu U/c = \dfrac{2\pi\mu r N}{c}$.

Friction force $F = \tau \times (2\pi r l)$ and torque $T = F\,r$:

$$T = \frac{4\pi^2 \mu r^3 l N}{c}$$

Also $T = f W r$ and $W = p\,(2r)\,l$ (projected area pressure). Therefore

$$f = \frac{T}{W r} = \frac{4\pi^2\mu r^3 l N}{c \cdot p\,(2rl)\, r} = 2\pi^2\,\frac{\mu N}{p}\,\frac{r}{c}$$

This is the Petroff equation.

Numerical

Data: $r = 37.5$ mm, $l = 75$ mm, $c = 0.075$ mm, $r/c = 500$, $N = 1000/60 = 16.67$ rev/s, $\mu = 0.025$ Pa.s, $W = 3000$ N.

$$p = \frac{W}{ld} = \frac{3000}{0.075 \times 0.075} = 0.533\times10^6\ \text{Pa}$$ $$\frac{\mu N}{p} = \frac{0.025 \times 16.67}{0.533\times10^6} = 7.81\times10^{-7}$$ $$f = 2\pi^2 \times 7.81\times10^{-7} \times 500 = 0.00771$$ $$\begin{aligned} T &= f W r = 0.00771 \times 3000 \times 0.0375 = 0.867\ \text{N.m} \\ \text{Power loss} &= T\omega = 0.867 \times 2\pi \times 16.67 = 90.8\ \text{W} \end{aligned}$$ $$S = \left(\frac{r}{c}\right)^2\frac{\mu N}{p} = 500^2 \times 7.81\times10^{-7} = 0.195$$

Answer: f = 0.0077; T = 0.867 N.m; power loss = 90.8 W; Sommerfeld number = 0.195.